Question

expand the function.
$f(x) = (2x - 3)^4$
$?x^4 + \square x^3 + \square x^2 + \square x + \square$

Explanation:

Step1: Recall Binomial Theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n!=n\times(n - 1)\times\cdots\times1\). For \((2x-3)^{4}\), we have \(a = 2x\), \(b=-3\), and \(n = 4\).

Step2: Calculate each term

• For \(k = 0\): \(\binom{4}{0}(2x)^{4}(-3)^{0}=1\times16x^{4}\times1 = 16x^{4}\)
• For \(k = 1\): \(\binom{4}{1}(2x)^{3}(-3)^{1}=\frac{4!}{1!3!}\times8x^{3}\times(-3)=4\times8x^{3}\times(-3)=-96x^{3}\)
• For \(k = 2\): \(\binom{4}{2}(2x)^{2}(-3)^{2}=\frac{4!}{2!2!}\times4x^{2}\times9 = 6\times4x^{2}\times9=216x^{2}\)
• For \(k = 3\): \(\binom{4}{3}(2x)^{1}(-3)^{3}=\frac{4!}{3!1!}\times2x\times(-27)=4\times2x\times(-27)=-216x\)
• For \(k = 4\): \(\binom{4}{4}(2x)^{0}(-3)^{4}=1\times1\times81 = 81\)

Answer:

The expanded form of \(f(x)=(2x - 3)^{4}\) is \(16x^{4}-96x^{3}+216x^{2}-216x + 81\). So the coefficients are \(16\) (for \(x^{4}\)), \(-96\) (for \(x^{3}\)), \(216\) (for \(x^{2}\)), \(-216\) (for \(x\)) and \(81\) (constant term).